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Proof step 32 from proof lookup-list-put-ex (conclusion)

(st[r0 -- sk, val][r -- sks]) ≠ st[r -- sks] ⊦ ∃ sks0, sks1. sks = sks0 + (sk + sks1) ∧ st[r -- sks0] .refval = r0

User applied rule structural induction

And obtained the following new goals

Proof step 1:
⊦ st[r -- []] = refval(r)
Proof step 31:
∀ st, r0, sk, val, r. (st[r0 -- sk, val][r -- sks0]) ≠ st[r -- sks0] → (∃ sks2, sks1. sks0 = sks2 + (sk + sks1) ∧ st[r -- sks2] .refval = r0), (st[r0 -- sk, val][r -- (sk0 + sks0)]) ≠ st[r -- (sk0 + sks0)] ⊦ ∃ sks2, sks1. sk0 + sks0 = sks2 + (sk + sks1) ∧ st[r -- sks2] .refval = r0

Rule argument

[VARWITHVARSEQSARG : sks, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : storekeys generated by [], +;], [CSIMPNAMEDSEQ : javavalue, , inj-03, ⊦ refval(r) = refval(r0) ↔ r = r0])]]